Chemistry · 2020
JEE Main · 9 January 2020, Shift 2 · Q27
The solubility product of Cr ( OH )_3 at 298 K is 6.0 × 10^-31. The concentration of hydroxide ions in a saturated solution of Cr ( OH )_3 will be:
The solubility product of $\displaystyle \mathrm{Cr}(\mathrm{OH})_3$ at $\displaystyle 298$ K is $\displaystyle 6.0 \times 10^{-31}$. The concentration of hydroxide ions in a saturated solution of $\displaystyle \mathrm{Cr}(\mathrm{OH})_3$ will be :
Official answer
(1)
$\displaystyle \left(18 \times 10^{-31}\right)^{1 / 4}$
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